# Special Relativity and Indices

LagrangeEuler
Can someone explain me difference between
$$A^{\mu}_{\hspace{0.2cm} \nu}$$
$$A^{\hspace{0.2cm} \mu}_{\nu}$$
and
$$A^{\mu}_{\nu}$$?

Gold Member
2022 Award
The first two expressions are correct and usually denote components of a 2nd-rank tensor. You can lower and raise indices with the metric components ##g_{\mu \nu}## and ##g^{\mu \nu}##, respectively, i.e., you have
$${A_{\nu}}^{\mu} = g_{\nu \sigma} g^{\mu \rho} {A^\sigma}_{\rho}.$$
The last expression should be avoided, because the horizontal placement of the indices is not indicated. It's ok if the tensor ##A## is symmetric, i.e., if ##A_{\mu \nu}=A_{\nu \mu}##, because (only!) then
$${A^{\mu}}_{\nu} = g^{\mu \rho} A_{\rho \nu} = g^{\mu \rho} A_{\nu \rho} ={A_{\nu}}^{\mu},$$
and the horizontal ordering is not important.

Dale, Orodruin, LagrangeEuler and 2 others
LagrangeEuler
The first two expressions are correct and usually denote components of a 2nd-rank tensor. You can lower and raise indices with the metric components ##g_{\mu \nu}## and ##g^{\mu \nu}##, respectively, i.e., you have
$${A_{\nu}}^{\mu} = g_{\nu \sigma} g^{\mu \rho} {A^\sigma}_{\rho}.$$
The last expression should be avoided, because the horizontal placement of the indices is not indicated. It's ok if the tensor ##A## is symmetric, i.e., if ##A_{\mu \nu}=A_{\nu \mu}##, because (only!) then
$${A^{\mu}}_{\nu} = g^{\mu \rho} A_{\rho \nu} = g^{\mu \rho} A_{\nu \rho} ={A_{\nu}}^{\mu},$$
and the horizontal ordering is not important.
Is there some connection with matrices? For instance, if we have two indices.
$$A^{\mu}_{\hspace{0.2cm}\nu}$$ what is the row and what is the column? And in this case
$$A^{\hspace{0.2cm}\mu}_{\nu}$$ what is the row and what is the column?

Staff Emeritus
Homework Helper
Gold Member
Tensors are not matrices and matrices are not tensors. Tensors (of rank 2) may be represented by matrices in some basis but it is then up to you to define how indices correspond to rows and columns in a consistent manner.

robphy, vanhees71 and cianfa72
cianfa72
And in this case $$A^{\hspace{0.2cm}\mu}_{\nu}$$ what is the row and what is the column?
Typically when in a given basis you represent a tensor as a matrix the first index (on the left) is the row and the second the column. So in your example ##\nu## is the row and ##\mu## the column.

Last edited:
vanhees71 and LagrangeEuler